Precalculus Examples

27 x^{4} - x

$27 x^{4} - x$

Step 1

Factor

x

$x$ out of

27 x^{4} - x

$27 x^{4} - x$ .

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Step 1.1

Factor

x

$x$ out of

27 x^{4}

$27 x^{4}$ .

x (27 x^{3}) - x

$x (27 x^{3}) - x$

Step 1.2

Factor

x

$x$ out of

- x

$- x$ .

x (27 x^{3}) + x \cdot - 1

$x (27 x^{3}) + x \cdot - 1$

Step 1.3

Factor

x

$x$ out of

x (27 x^{3}) + x \cdot - 1

$x (27 x^{3}) + x \cdot - 1$ .

x (27 x^{3} - 1)

$x (27 x^{3} - 1)$

x (27 x^{3} - 1)

$x (27 x^{3} - 1)$

Step 2

Rewrite

27 x^{3}

$27 x^{3}$ as

{(3 x)}^{3}

${(3 x)}^{3}$ .

x ({(3 x)}^{3} - 1)

$x ({(3 x)}^{3} - 1)$

Step 3

Rewrite

1

$1$ as

1^{3}

$1^{3}$ .

x ({(3 x)}^{3} - 1^{3})

$x ({(3 x)}^{3} - 1^{3})$

Step 4

Since both terms are perfect cubes, factor using the difference of cubes formula,

a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})

$a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where

a = 3 x

$a = 3 x$ and

b = 1

$b = 1$ .

x ((3 x - 1) ({(3 x)}^{2} + 3 x \cdot 1 + 1^{2}))

$x ((3 x - 1) ({(3 x)}^{2} + 3 x \cdot 1 + 1^{2}))$

Step 5

Factor.

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Step 5.1

Simplify.

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Step 5.1.1

Apply the product rule to

3 x

$3 x$ .

x ((3 x - 1) (3^{2} x^{2} + 3 x \cdot 1 + 1^{2}))

$x ((3 x - 1) (3^{2} x^{2} + 3 x \cdot 1 + 1^{2}))$

Step 5.1.2

Raise

3

$3$ to the power of

2

$2$ .

x ((3 x - 1) (9 x^{2} + 3 x \cdot 1 + 1^{2}))

$x ((3 x - 1) (9 x^{2} + 3 x \cdot 1 + 1^{2}))$

Step 5.1.3

Multiply

3

$3$ by

1

$1$ .

x ((3 x - 1) (9 x^{2} + 3 x + 1^{2}))

$x ((3 x - 1) (9 x^{2} + 3 x + 1^{2}))$

Step 5.1.4

One to any power is one.

x ((3 x - 1) (9 x^{2} + 3 x + 1))

$x ((3 x - 1) (9 x^{2} + 3 x + 1))$

x ((3 x - 1) (9 x^{2} + 3 x + 1))

$x ((3 x - 1) (9 x^{2} + 3 x + 1))$

Step 5.2

Remove unnecessary parentheses.

x (3 x - 1) (9 x^{2} + 3 x + 1)

$x (3 x - 1) (9 x^{2} + 3 x + 1)$

x (3 x - 1) (9 x^{2} + 3 x + 1)

$x (3 x - 1) (9 x^{2} + 3 x + 1)$

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